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hram777 [196]
3 years ago
9

Need help ASAP! I will give Brainliest answer to the person who answers best!

Mathematics
1 answer:
tatiyna3 years ago
3 0

First find the missing length of the left triangle with the pythagorean theorem a² + b²= c² → 4² + 6² = c² → 16 + 36 = c² → 52 = c² → √52 = c

Now if you turn your head you can see a right triangle made of the two smaller triangles. The √52 is one of the legs. We know that the hypotenuse is 6 + y so we can substitute these values into the pythaorean theorem as well to find the blank side length of the smaller triangle to the right.

√52² + b² = (6 + y)² → 52 + b² = (6 + y)(6 + y) I know this seems like a lot but this is how I'd do it. You need to multiply (foil method) the (6+y)(6+y) to get 36 + 6y + 6y + y² which simplifies to 36 + 12y + y². Here's what we have: to find the hypotenuse of the smaller triangle to the right we need to simplify 52 + b² = 36 + 12y + y²; subtract 52 from both sides

b² = y² + 12y - 16 (I rearranged to make it easier to see); square root both sides

b = \sqrt{y²  + 12y - 16} this is the hypotenuse of the smaller triangle to the right

Then we can use the pythagorean theorem for the small triangle to the right to eventually find the value of y.

4² + y² = \sqrt{y²}+12y - 16 } ²

16 + y² = y² + 12y - 16; now solve, add 16 to both sides

32 + y² = y² + 12y; subtract y² from both sides

32 = 12y; divide both sides by 12

approximately 2.6 = y but since it says do not approximate, I would put y = 32/12

I hope that helps!!!

Ignore the weird A under the radical, I don't know why that's there. It's supposed to be just y²

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3 years ago
He dimensions are tripled. The new surface area would be times larger than the original surface area.
ddd [48]

Answer:

The new surface area would be 9 times larger than the original surface area.

Step-by-step explanation:

Here we do not know what was the original shape, but we will see that it does not matter.

Let's start with a square of side length L.

The original area of this square will be:

A = L^2

Now if each dimension is tripled, then all the sides of the square now will be equal to 3*L

Then the new area of the square is:

A' = (3*L)^2 = (3*L)*(3*L) = 9*L^2 = 9*A

So the new surface area is 9 times the original one.

Now, if the figure was a circle instead of a square?

For a circle of radius R, the area is:

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Now if the dimensions of the circle are tripled, the new radius will be 3*R

Then the new area of the circle is:

A' = pi*(3*R)^2 = pi*9*R^2 = 9*(pi*R^2) = 9*A

Again, the new area is 9 times the original one.

If the figure is a triangle?

We know that for a triangle of base B and height H, the area is:

A = B*H/2

If we triple each measure, we will have a base 3*B and a height 3*H

Then the new area is:

A' = (3*B)*(3*H)/2 = (3*3)*(B*H/2) = 9*(B*H/2) = 9*A

Again, the new area is 9 times the original area.

So we can conclude that for any shape, the new area will be 9 times the original area.

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3 years ago
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